Visible Light Reflection from Oil-Water Interface in Parking Lot

Click For Summary
SUMMARY

The discussion focuses on calculating the wavelengths of visible light that will strongly reflect off an oil-water interface, specifically with oil having a refractive index of 1.25 and water with a refractive index of 1.33. The oil layer thickness is 242 nm. To find the wavelengths that constructively interfere, the formula 2t = m * λ/n2 is utilized, where t is the thickness of the oil, and λ is the wavelength of light. Participants emphasize starting with m = 1 and calculating λ for each integer value of m until reaching the visible spectrum range of 400 nm to 700 nm.

PREREQUISITES
  • Understanding of refractive indices, specifically for oil (1.25) and water (1.33).
  • Familiarity with the concept of constructive interference in optics.
  • Knowledge of the visible light spectrum, particularly the wavelength range of 400 nm to 700 nm.
  • Proficiency in using the formula 2t = m * λ/n2 for optical calculations.
NEXT STEPS
  • Calculate the wavelengths of visible light using the formula 2t = m * λ/n2 for m values starting from 1.
  • Explore the effects of varying oil thickness on the colors reflected at the oil-water interface.
  • Investigate the principles of thin film interference in optics.
  • Learn about the applications of oil-water interfaces in real-world scenarios, such as environmental science and material science.
USEFUL FOR

Students and professionals in physics, particularly those studying optics, as well as environmental scientists and engineers working with oil-water interactions.

truckerron1
Messages
8
Reaction score
0
A tiny layer of oil (n = 1.25) is situated on top of a water puddle (n = 1.33) in a parking lot. If the thickness of the oil is 242 nm, the what color(s) of visible light will give a strong reflection?

2t=m*lambda/n2

2t=242nm/1.25 and then 2t=242nm/1.33
i just don't know what goes where on this problem please help thanks ron
 
Physics news on Phys.org
242 nm is the thickness of the oil (t), not the wavelength of the light (lambda).
 
so how do i set it up if t=242
 
so would it be
2(242)=m*1.25/1.33
 
No, NONE of the light that gets reflected travels through the water.
The extra path length (which is responsible for the constructive interference)
travels down thru the OIL and back up thru the OIL.
You should expect the OIL thickness to be associated with the OIL index.

You're trying to SOLVE for lambda ...
 
Last edited:
In addition to lightgrav's useful comment, I'll add that at this point you may be wondering what to use for m?

Each integer value of m (1, 2, 3, ...) gives a possible wavelength, in principle. However, you specifically need only the wavelengths that are visible, that is, in the range 400nm to 700nm. So start with m = 1 and calculate lambda for each value of m in turn. It will be pretty obvious when you can safely stop (i.e. after you've gone through the entire visible range).
 

Similar threads

Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
5
Views
8K
Replies
1
Views
3K
Replies
4
Views
8K
Replies
3
Views
7K